In 2013, a study found that U. S. consumers consumed soft drinks with a mean of 42.2 gallons per year and variance of 169. Ninety-five percent of consumers in the U. S. consumed less than how many gallons?

Round to two decimals and include a leading zero if necessary.

Ninety-five percent of consumers in the U.S. consumed less than 63.59 gallons.

Step-by-step explanation:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean and standard deviation , the zscore of a measure X is given by:

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

The standard deviation is the square root of the variance, so

Also, the mean is 42.2, so

Ninety-five percent of consumers in the U.S. consumed less than how many gallons?

The 95th percentile, which is the value of X when Z has a pvalue of 0.95. So X when

Ninety-five percent of consumers in the U.S. consumed less than 63.59 gallons.

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